A box contains $5$ red marbles, $4$ green marbles, and $5$ blue marbles. If we choose a marble, then another marble without putting the first one back in the box, what is the probability that the first marble will be red and the second will be red as well? Write your answer as a simplified fraction.
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a red marble and leaving it out. Event B is picking another red marble. Let's take the events one at at time. What is the probability that the first marble chosen will be red? There are $5$ red marbles, and $14$ total, so the probability we will pick a red marble is $\dfrac{5} {14}$ After we take out the first marble, we don't put it back in, so there are only $13$ marbles left. Also, we've taken out one of the red marbles, so there are only $4$ left altogether. So, the probability of picking another red marble after taking out a red marble is $\dfrac{4} {13}$ Therefore, the probability of picking a red marble, then another red marble is $\dfrac{5}{14} \cdot \dfrac{4}{13} = \dfrac{10}{91}$